I never rely on Pascal's triangle or any other memorized shortcut. I always do the algebra out in long hand. I find it more informative and quite relaxing as well. I believe that the joy of mathematics doesn't come from getting a quick answer but rather in knowing and logically following all of the rules for the type(s) of math objects you are working with. The beauty of solving any math problem is producing a logical workflow that can be read as a rigorous proof by anyone, not an exercise for the reader to figure out on their own. My two cents worth.
How is it more informative to not use pascals triangle? It literally is the fundamental underlying structure behind binomial expansion to any degree and it is literally key to approximating roots, etc. I understand that it feels more rewarding to expand correctly and it does help with algebraic multiplication for higher degrees. But i feel that skill is best developed through just general mathsmatical practice. Just my take anyway, if you feel it helps you best go ahead. im just really curious.
Totally loved your way, Another way - We know if x+y+z=0 then X^3+ y^3 +z^3= 3xyz (it is what it is, search) Now from the first equation that's the sum of cuberoots pf x ,y and z= 0 We can substitute in second equation that is (x+y+z)^3= 3yz As (3 multiplied by cube root of x, y and z) ^3 = 3yz Now 27xyz= 3yz And x=1/9 And the others solution as you did But in the solution I gave you just know a simple formula and apply it which saves time but does not develop approach.
I don’t remember when I subscribed to you (probably when I was in school) but I very happy I still am. This was very relaxing and more productive than doom scrolling
An excellent problem and solution. I am having visions of an xyz-coordinate system with a yz-plane at x=1/9, crisscrossed by two lines passing through the origin. No, that's not correct, is it? To be explicit: When x=1/9, don't we still have to solve for y and z? (Okay, I guess the problem didn't ask for that.) Even more impressive is that you got through the entire lesson with once saying, "zed"!
At first, I had the same thought process you did to get to x=1/9, but I made much more work for myself than was necessary by missing some substitutions that are so blatantly obvious in hindsight; and now I have a headache. That was fun. Lol
But cubic root is not the same thing as 1/3 power. In cubic root the argument can be any real number, even negative, but if we use power notation, the argument must be positive.
BTW I noticed, that the 4th option is actually a common particular case of 2 above. they state if y or z =0, them x = - (another letter), but if it's also 0, than equal to -0 = 0 LoL
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You're a fantastic math communicator.
I never rely on Pascal's triangle or any other memorized shortcut. I always do the algebra out in long hand. I find it more informative and quite relaxing as well. I believe that the joy of mathematics doesn't come from getting a quick answer but rather in knowing and logically following all of the rules for the type(s) of math objects you are working with. The beauty of solving any math problem is producing a logical workflow that can be read as a rigorous proof by anyone, not an exercise for the reader to figure out on their own. My two cents worth.
I mean I don’t agree, but good to know your opinion
How is it more informative to not use pascals triangle? It literally is the fundamental underlying structure behind binomial expansion to any degree and it is literally key to approximating roots, etc. I understand that it feels more rewarding to expand correctly and it does help with algebraic multiplication for higher degrees. But i feel that skill is best developed through just general mathsmatical practice. Just my take anyway, if you feel it helps you best go ahead. im just really curious.
Beautiful, detailed, and clear explanation! And, I must say, beautiful handwriting. Keep up good work!
Totally loved your way,
Another way -
We know if x+y+z=0 then
X^3+ y^3 +z^3= 3xyz (it is what it is, search)
Now from the first equation that's the sum of cuberoots pf x ,y and z= 0
We can substitute in second equation that is (x+y+z)^3= 3yz
As
(3 multiplied by cube root of x, y and z) ^3 = 3yz
Now 27xyz= 3yz
And x=1/9
And the others solution as you did
But in the solution I gave you just know a simple formula and apply it which saves time but does not develop approach.
I really enjoyed this problem...... thank you for providing this..... great work ❤❤❤
Love from India ❤❤❤
great work man!
love from Brazil
A very beautifull solution! Your videos are asome! Greetings from Paraguay
actually last case is already covered.
I don’t remember when I subscribed to you (probably when I was in school) but I very happy I still am. This was very relaxing and more productive than doom scrolling
Glad you are still here.
An excellent problem and solution. I am having visions of an xyz-coordinate system with a yz-plane at x=1/9, crisscrossed by two lines passing through the origin. No, that's not correct, is it? To be explicit: When x=1/9, don't we still have to solve for y and z? (Okay, I guess the problem didn't ask for that.)
Even more impressive is that you got through the entire lesson with once saying, "zed"!
Did I really say 'zed'? 🤣🤣🤣🤣
Yes brother
@@PrimeNewtonsZed is correct. Don't let the Americans influence you. Everyone else says zed.
Very small mistake at the end, on 3rd part of the board x =1/9 (x,y != 0) should be (y,z != 0)
Nothing major very good video :)
Thank yoy!
My brain took a break there 🙃 😪
@@PrimeNewtons Don't worry it happens to every one
@@PrimeNewtonsbrother where are you from
@@PrimeNewtons Since you didn't you give more information about X,Y and Z , I immediately said that they're all equal to zero😅
At first, I had the same thought process you did to get to x=1/9, but I made much more work for myself than was necessary by missing some substitutions that are so blatantly obvious in hindsight; and now I have a headache. That was fun. Lol
Otro excelente video
11:00 but if x = -z, the cube root of a negative should be imaginary right? how do we get 0 from positive + imaginary?
負數的立方根可以不是虛數
Let b = a³ (a
The last codition is not required because from the second and third condition, we can get x=0 if y=z=0
(a+b)^3=a^3+b^3+3ab(a+b)
This is the key to the cubic equation solving
But cubic root is not the same thing as 1/3 power. In cubic root the argument can be any real number, even negative, but if we use power notation, the argument must be positive.
Not an important issue: at 9.40 Case 1 y,z are not equal 0. As mentioned 1 minute before.
BTW I noticed, that the 4th option is actually a common particular case of 2 above. they state if y or z =0, them x = - (another letter), but if it's also 0, than equal to -0 = 0 LoL
Thanks for an other video master
7:30 the face is so funny
Prime Newtons leads the way! 🎉😊
@PrimeNewtons Since you didn't you give more information about X,Y and Z , I immediately said that they're all equal to zero
y, z also can be solved.
Misheard cube root as cubert (Q-bert).
Added to my mathematical vocabulary. : )
Another answer is x=3^1/3 /9
please solve this f(x)_f'(×)=x^2
I don't understand the equation. Please write on paper and email a picture.
How about x,y,z=0
Wee